Circuit Theory Homework Question Answer 2022

Running head: CIRCUIT THEORY 1
Circuit Theory
Name
Institution

CIRCUIT THEORY 2
Circuit Theory Homework
Question 1
Part a (Thevenin Theorem)
V 1= √2 × 415cos ( 100 πt ) , V 1 ( RMS )=415 ∠90 °
V 2= √2 × 415sin ( 100 πt ) , V 2 ( RMS )=415∠ 0 °
We use the RMS values of V 1 and V 2
We remove the load and short the source voltages to obtain the circuit shown below.
Then apply the KCL node ZTh and calculate the Thevenin equivalent voltage V Th
V Th−V 1
j 4 + V Th−V 2
j 6 =0
V Th−V 1
j 4 = V 2−V Th
j6
( V Th−V 1 ) j 6= ( V 2−V Th ) j 4
V Th j 6−V 1 j6=V 2 j 4−V Th j 4
V Th j 6+V Th j 4=V 2 j 4 +V 1 j 6
V Th( j 6+ j 4)=(V 2 j 4 +V 1 j6)
V Th =(V 2 j 4+V 1 j 6)
j 6+ j 4 = 415 ∠0 ° ( 4 ∠ 90 ° ) +415 ∠ 90 ° (6 ∠90 ° )
10 ∠ 90 ° =299.2608∠ 56.31°
ZTh= j4 Ω∨¿ j6 Ω= j 4 × j6
j 4 + j 6 = j 2.4=2.4 ∠ 90° Ω

CIRCUIT THEORY 3
The Thevenin equivalent circuit becomes:
ZLoad =50 ohms at 0.7 pf lag
φ =cos−1 0.7=45.57 °
ZLoad =50∠ 45.57 ° Ω
iTh= V Th
ZTh + Z Load
= 299.2608∠ 56.31°
2.4 ∠90 °+50 ∠ 45.57 ° =5.7838∠ 8.88 °
Part b (Superposition Theorem)
First, we remove V 2 and replace it with a short circuit then calculate V x using the voltage divider
rule as shown below.
V x= V 1
j 4+ j6∨¿ ZLoad
× j 6∨¿ ZLoad = 415 ∠90 °
4 ∠ 90 ° + 6 ∠ 90 ° × 50 ∠ 45.57°
6 ∠ 90°+50 ∠ 45.57 °
× 6 ∠ 90 ° × 50∠ 45.57 °
6 ∠90 °+50 ∠ 45.57 °
¿ 415∠ 90 °
4 ∠ 90 °+5.51∠ 85.58 ° ×5.51 ∠85.58 °
¿ 240.6209 ∠8 8.139 ° V

CIRCUIT THEORY 4
i' = V x
ZLoad
=240.6209 ∠ 88.139° V
50 ∠ 45.57° =4.8124 ∠ 42.5692 ° A
Second, we remove V 1 and replace it with a short circuit then calculate V y using the voltage
divider rule as shown below.
V y= V 2
j6+ j 4∨¿ Z Load
× j 4∨¿ Z Load= 415 ∠ 0 °
6 ∠ 90 °+ 4 ∠ 90° ×50 ∠ 45.57 °
4 ∠ 90 °+50 ∠ 45.57°
× 4 ∠ 90 ° × 50∠ 45.57 °
4 ∠ 90°+50 ∠ 45.57 °
¿ 415∠ 0 °
6 ∠90 °+3.7785 ∠86.97 ° × 3.7 785 ∠86.97 °
¿ 160.4139 ∠−1.86 °V
i' '= V y
ZLoad
= 160.4139∠−1.86 ° V
50 ∠ 45.57 ° =3.2083 ∠−47 .43° A
By superposition, the total current is:
I =i'+i' '=4.8124 ∠ 42.5692° A +3.2083∠−47.43 ° A=5.7838 ∠ 8.88° A
Part c (Norton’s Theorem)
First, we short circuit load and calculate the Norton current provided by both the voltage sources
V 1 and V 2 as follows:
I 1=V 1 (RMS )
X1
= 415 ∠ 90 °
4 ∠ 90 ° =103.75∠ 0 ° A
I 2=V 2(RMS )
X2
= 415∠ 0 °
6 ∠ 90 ° =69.1667 ∠−9 0 ° A